The magnitude of the resulting vector (i.e., speed) can be calculated trivially since their movement is perpendicular on a plane, as the root of sum of squares, which many could recognize as the Pythagorean theorem:

√((5 ft/s)² + (1 ft/s)²) = √26 ft/s ≈ 5.1 ft/s

You can verify this by finding that their average speed apart is the same at all times (for all t > 0):

Vavg = √((t * 5 ft/s)² + (t * 1 ft/s)²) / t = √(t² * ((5 ft/s)² + (1 ft/s)²)) / t = √26 ft/s

[–] [email protected] 24 points 3 months ago* (last edited 3 months ago) (1 children)

Don't forget to calculate the location where everything about them began and then include the curvature of Earth considering the latitude of said location into your speed calculation.

[–] SpaceNoodle 34 points 3 months ago (3 children)

No, they're spherical children in a vacuum.

[–] answersplease77 13 points 3 months ago

for approximation we can assume that the boy is a point mass and the girl is a lie

[–] [email protected] 5 points 3 months ago

Oh, so we have to calculate the gravitational attraction pulling them back. Fucking hell

[–] Hamartia 2 points 3 months ago

Augustus! Save some room for later.

[–] Randelung 3 points 3 months ago* (last edited 3 months ago) (1 children)

https://en.m.wikipedia.org/wiki/Spherical_geometry

I couldn't find 'potatoy geometry' for a better approximation of earth.

[–] SpaceNoodle 1 points 3 months ago (2 children)

You'll note that I already assumed that they were on a plane, not the surface of a sphere.

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[–] De_Narm 41 points 3 months ago* (last edited 3 months ago) (1 children)

It's been a while, but I think it's quite trivial.

After one second, they span a right angled triangle, therefore (using a² + b² = c²) their distance is √(5²+1²) = ~5.1 ft

They move at constant speed, therefore they seperate at 5.1 ft/s. That means at 5s it's just 5.1 × 5 = 25.5 ft for the distance and their speed is still the same.

[–] [email protected] 4 points 3 months ago* (last edited 3 months ago) (2 children)

~~They each move at a constant speed, but the distance between them doesn't increase at a constant pace. See my other comment.~~

Edit: I am dumb, and looked at the wrong number.

[–] De_Narm 12 points 3 months ago* (last edited 3 months ago) (1 children)

I'm trying to apply the most simple math possible and it seems to add up.

After one second, their distance is √(5² + 1²) = ~5.1 ft

After two seconds, their distance is √(10² + 2²) = ~10.2 ft

After three seconds, it's √(15² + 3²) = ~15.3 ft

As speed is the rate of change of distance over time, you can see it's a constant 5.1 ft/s. You're free to point out any error, but I don't think you need anything more than Pythagoras' theorem.

The question specifically asks for their seperation speed at 5s to ignore any initial change in their speed as they first need to accelerate, I'd assume.

[–] [email protected] 7 points 3 months ago (1 children)

Ah sorry, I'm tired and made a mistake. I quickly made a spreadsheet (because keeping track of numbers is hard), and I was looking at the wrong column in the sheet. My bad!

[–] [email protected] 8 points 3 months ago (1 children)

You were tired so you made a spreadsheet to calculate the differential equation quiz from a meme?

[–] [email protected] 3 points 3 months ago

Yes, compared to doing the calculations in my head lol

I work in mysterious ways

[–] [email protected] 7 points 3 months ago

I don't see why the distance between them isn't growing at a constant speed.

At any given time t seconds after separation, the boy is 5t north, and the girl is 1t east. The distance between them is defined by the square root of ((5t)^2 + (t)^2 ), or about 5.099t.

In other words, the distance between them is simply a function defined as 5.099t, whose first derivative with respect to time is just 5.099.

[–] [email protected] 39 points 3 months ago (1 children)

Depends on where they met each other. If they for example fell in love during the main event of a trip to the north pole, that would change things a lot.

[–] [email protected] 18 points 3 months ago (2 children)

there is no north at the north pole so actually that's the one place it can't be

[–] [email protected] 13 points 3 months ago

If you're at the south pole, would every direction count as north?

[–] [email protected] 7 points 3 months ago

Sure, but there is a north say 30 ft away from the north pole.

[–] Missmuffet 35 points 3 months ago (1 children)

Its pretty convenient that its raining, which means you can ignore the coefficient of friction since the surface is slippery

[–] itsnotits 4 points 3 months ago

It's* pretty convenient that it's* raining

[–] [email protected] 11 points 3 months ago

It doesn't matter what the actual answer is; to both the boy and the girl it feels like C.

[–] [email protected] 9 points 3 months ago

reminds me of that one song, proof that geometric construction can solve all love affairs or something like that

[–] [email protected] 6 points 3 months ago (1 children)

what

[–] someguy3 12 points 3 months ago

Who hurt the math teacher?

[–] [email protected] 3 points 3 months ago (1 children)

9,14 Meter

[–] [email protected] 3 points 3 months ago* (last edited 3 months ago) (1 children)

The question states "how fast", not "how far", thus you need to give the acceleration at that moment.

At t=0, the boy and girl both haven't moved, so their positions are 0. The distance between them is also 0, as is their acceleration.

The boy's distance in meters is t*1.524, the girl's distance is t*0.3048. The distance between them is sqrt( b^2 * g^2 ). The velocity is the current distance minus the previous distance.

At t=1, b=1.524m, g=0.305, d=sqrt( g^2 * g^2 )=0.465, v=d-d^(t-1)=0.465m/s.

At t=5, b=7.62, g=1.524, d=11.613, and v=4.181m/s.

Edit: fixed markdown

[–] SpaceNoodle 4 points 3 months ago (1 children)

Velocity is not the difference between distances.

[–] [email protected] 3 points 3 months ago (1 children)

It’s the difference of distances apart over time. Aka how fast bf is moving away from gf, aka what the question is asking for.

Yes, if you want to be pedantic, velocity a vector with direction, so I guess you’d have to frame the question relative to either the boyfriend or girlfriend, but I don’t think the difference between speed and velocity is part of the question.

[–] SpaceNoodle 2 points 3 months ago (2 children)

Speed is just the magnitude of velocity.

My point is that OC was completely missing the mark by not properly accounting for time.

[–] [email protected] 2 points 3 months ago (1 children)

Hi, I made this in 5 mins because I was bored, but it's late and I'm tired, so could you please explain what I would have to fix in my comment?

[–] [email protected] 3 points 3 months ago (1 children)

You want to figure out distance per second. One way to do this is calculate distance apart at t=0,1,2…

The difference between each point would be the average speed over that second.

Using sqrt(b^2+g^2):

t0 = 0 t1 = 1.554m
s1 = (1.554m-0m)/1s = 1.554m/s t2 = 3.108m
s2=(3.108m-1.554m)= 1.554m/s

As you continue this you will see they travel at a constant speed apart from each other. The reason this is working is because you need to divide distance by time. Dividing by 1 second won’t change the value of the number after you subtract. If you notice you can do (t2-t0)/2s and also get the same answer.

[–] [email protected] 1 points 3 months ago

Ahhh okay, thanks

[–] [email protected] 1 points 3 months ago (4 children)

My mistake, I didn’t check his math. I thought he was saying if you take distance apart at t(n) and subtract distance apart at t(n-1) you will get distance/sec.

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[–] Etterra 2 points 3 months ago

7 Hallmarks per Homecoming.

[–] SatansMaggotyCumFart 1 points 3 months ago (1 children)

The answer is six feet per second, right?

[–] FierySpectre 7 points 3 months ago (2 children)

Can't tell if that's a genuine answer or a joke sooo...

They're moving perpendicular so unfortunately it's not that easy... I'll leave the calculations to someone who didn't fail math though

[–] SatansMaggotyCumFart 3 points 3 months ago (1 children)

Oh that makes sense why the elapsed time would factor into the equation too.

[–] Gordon 3 points 3 months ago (4 children)

To trick you.

Now if the question had been something like:

the boy starts out running at a speed of 5ft/s then slows to 2.5ft/s over the next 5 seconds...

Then you'd need differential equations or whatever the class is, and you'd have to use more than just basic trig, but as written, it's just basic trig and the time is irrelevant.

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[–] coffee_whatever 2 points 3 months ago (1 children)

Wouldn't that be just a right triangle? One side 5×5=25 feet long, the other 5×1=5 feet long so the hypotenuse would be the square root of (25²+5²) or around 25,5 feet.

That's basic geometry right? It's been quite a few month but I'm pretty sure I still can do highschool level maths.

[–] FierySpectre 2 points 3 months ago (1 children)

That's the distance yes, but how fast are they moving away from eachother

[–] Gordon 3 points 3 months ago (1 children)

(5^2 + 1^2)^0.5 = 5.1ft/s

[–] FierySpectre 2 points 3 months ago* (last edited 3 months ago)

Well, like I said I failed math so I'll take your word for it

Edit: autocorrect fucked me

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